AS October 2020 Q1
1. A plumbing company receives call-outs during the working day at an average rate of 2.4 per hour.
The company has enough staff to respond to 28 call-outs in an 8-hour working day.
In a random sample of 100 working days each of 8 hours,
| Scheme | Marks | AO |
|---|---|---|
| \(X \sim \mathrm{Po}(7.2)\) | M1 | 3.4 |
| \(\mathrm{P}(X = 7) = 0.14858\ldots\) awrt 0.149 | A1 | 1.1b |
| (2) |
Notes
M1: Writing or using \(\mathrm{Po}(7.2)\)
A1: awrt 0.149
| Scheme | Marks | AO |
|---|---|---|
| \(Y \sim \mathrm{Po}(19.2)\) | M1 | 3.3 |
| [\(\mathrm{P}(Y \gt 28) =\)] \(1 - \mathrm{P}(Y \leqslant 28) = 1 - 0.9780\ldots = 0.02199\ldots\) 0.022* | A1* | 1.1b |
| (2) |
Notes
M1: Writing or using \(\mathrm{Po}(19.2)\)
A1*: cso given answer with correct probability statement (e.g. \(1 - \mathrm{P}(Y \leqslant 28)\) ) and no incorrect working seen
| Scheme | Marks | AO |
|---|---|---|
| (i) [\(100 \times 0.022\)] awrt 2.2 | B1 | 1.1b |
| (1) | ||
| (ii) \(\sqrt{100(0.022)(1 - 0.022)}\) | M1 | 1.1b |
| \(= 1.466\ldots\) awrt 1.47 | A1 | 1.1b |
| (2) | ||
| (iii) \(\mathrm{B}(100, 0.022) \rightarrow \mathrm{Po}(2.2)\) | M1 | 3.4 |
| \(\mathrm{P}(W \geqslant 6) = 1 - \mathrm{P}(W \leqslant 5)\) [\(= 1 - 0.9750\ldots\)] | M1 | 1.1b |
| \(= 0.02490\ldots\) awrt 0.0249 | A1 | 1.1b |
| (3) | ||
| (10 marks) |
Notes
(c)(i) B1: awrt 2.2 (isw once awrt 2.2 is seen)
(c)(ii) M1: Correct expression including square root
A1: awrt 1.47
Watch out for \(\sqrt{2.2} = 1.483\ldots\) which is M0A0
(iii) M1: Approximating binomial (100, 0.022) with \(\mathrm{Po}(2.2)\) [may be seen in (i) or (ii)]
M1: Using \(1 - \mathrm{P}(W \leqslant 5)\) from Poisson distribution
A1: awrt 0.0249
Note: Using Binomial distribution \(1 - 0.9765588\ldots = 0.02344\ldots\) scores M0M0A0